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borishaifa [10]
3 years ago
9

Which system of equations has the same solution as the system below?

Mathematics
1 answer:
worty [1.4K]3 years ago
8 0

Answer: Both A, and C

Step-by-step explanation:

The answer to the first system of equations (2x+2y=16) would be

x=3 and y=5 ( 3x-y=4 )

Which means we have to find out which of the other equations has an x value of 3, and a y value of 5.

If A is 2x+2y=16, then x=3 and y=5

           6x-2y=8

If B is x+y=16, then x=5 and y=11

         3x-y=4

If C is 2x+2y=16, then x=3 and y=5

           6x-2y=8

If D is 6x+6y=48 , then x=-2 and y=10

            6x+2y=8

Both A and C are equal to the first system of equations, which means they are both correct answers.

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Find the equation of the directrix of the parabola x2=+/- 12y and y2=+/- 12x
PIT_PIT [208]

Answer:

  1. x^2 = 12 y equation of the directrix y=-3
  2. x^2 = -12 y equation of directrix y= 3
  3. y^2 = 12 x   equation of directrix x=-3
  4. y^2 = -12 x equation of directrix x= 3

Step-by-step explanation:

To find the equation of directrix of the parabola, we need to identify the axis of the parabola i.e, parabola lies in x-axis or y-axis.

We have 4 parts in this question i.e.

  1. x^2 = 12 y
  2. x^2 = -12 y
  3. y^2 = 12 x
  4. y^2 = -12 x

For each part the value of directrix will be different.

For x²  = 12 y

The above equation involves x² , the axis will be y-axis

The formula used to find directrix will be: y = -a

So, we need to find the value of a.

The general form of equation for y-axis parabola having positive co-efficient is:

x² = 4ay  eq(i)

and our equation in question is: x² = 12y eq(ii)

By putting value of x² of eq(i) into eq(ii) and solving:

4ay = 12y

a= 12y/4y

a= 3

Putting value of a in equation of directrix: y = -a => y= -3

The equation of the directrix of the parabola x²= 12y is y = -3

For x²  = -12 y

The above equation involves x² , the axis will be y-axis

The formula used to find directrix will be: y = a

So, we need to find the value of a.

The general form of equation for y-axis parabola having negative co-efficient is:

x² = -4ay  eq(i)

and our equation in question is: x² = -12y eq(ii)

By putting value of x² of eq(i) into eq(ii) and solving:

-4ay = -12y

a= -12y/-4y

a= 3

Putting value of a in equation of directrix: y = a => y= 3

The equation of the directrix of the parabola x²= -12y is y = 3

For y²  = 12 x

The above equation involves y² , the axis will be x-axis

The formula used to find directrix will be: x = -a

So, we need to find the value of a.

The general form of equation for x-axis parabola having positive co-efficient is:

y² = 4ax  eq(i)

and our equation in question is: y² = 12x eq(ii)

By putting value of y² of eq(i) into eq(ii) and solving:

4ax = 12x

a= 12x/4x

a= 3

Putting value of a in equation of directrix: x = -a => x= -3

The equation of the directrix of the parabola y²= 12x is x = -3

For y²  = -12 x

The above equation involves y² , the axis will be x-axis

The formula used to find directrix will be: x = a

So, we need to find the value of a.

The general form of equation for x-axis parabola having negative co-efficient is:

y² = -4ax  eq(i)

and our equation in question is: y² = -12x eq(ii)

By putting value of y² of eq(i) into eq(ii) and solving:

-4ax = -12x

a= -12x/-4x

a= 3

Putting value of a in equation of directrix: x = a => x= 3

The equation of the directrix of the parabola y²= -12x is x = 3

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